NCERT Class 11 Mathematics — Straight Lines, Miscellaneous Exercise on Chapter 9. All 23 questions solved.

This exercise combines everything from the chapter. Four techniques recur:

• Concurrency — find where two lines meet, then force the third through that point (questions 8, 9, 11, 23).
• Reflection — an image in a mirror line is found from two conditions: the midpoint lies on the line, and the joining segment is perpendicular to it (questions 17, 21).
• The parametric form of a line — a point at distance $r$ from $(x_1, y_1)$ in the direction $\theta$ is $\left(x_1 + r\cos\theta,\ y_1 + r\sin\theta\right)$ (question 15).
• Case-splitting on a modulus — the angle formula and the distance formula both carry one, and each sign gives a different answer (questions 10, 18).

Key insight. “Reach the path in the least time” (question 23) means travel the shortest distance, which means going along the perpendicular. Word problems in this exercise are geometry statements in disguise, and translating the phrase into the right geometric object is the whole difficulty.

Question 1

Find the values of $k$ for which the line $(k-3)x – \left(4 – k^2\right)y + k^2 – 7k + 6 = 0$ is

(a) parallel to the $x$-axis, (b) parallel to the $y$-axis, (c) passing through the origin.

Solution. Compare with $\mathrm Ax + \mathrm By + \mathrm C = 0$, where $\mathrm A = k – 3$, $\mathrm B = -(4 – k^2)$ and $\mathrm C = k^2 – 7k + 6$.

(a) Parallel to the $x$-axis means the equation has no $x$ term, so $\mathrm A = 0$ while $\mathrm B \ne 0$:

$$k – 3 = 0 \quad\Longrightarrow\quad k = 3$$

Check: at $k = 3$, $\mathrm B = -(4-9) = 5 \ne 0$ ✓

(b) Parallel to the $y$-axis means no $y$ term, so $\mathrm B = 0$:

$$4 – k^2 = 0 \quad\Longrightarrow\quad k = \pm2$$

At $k = 2$, $\mathrm A = -1 \ne 0$; at $k = -2$, $\mathrm A = -5 \ne 0$ ✓

(c) Passing through the origin means $(0,0)$ satisfies the equation, so the constant term vanishes:

$$k^2 – 7k + 6 = 0 \quad\Longrightarrow\quad (k-1)(k-6) = 0 \quad\Longrightarrow\quad k = 1 \text{ or } k = 6$$

Question 2

Find the equations of the lines which cut off intercepts on the axes whose sum and product are $1$ and $-6$ respectively.

Solution. With intercepts $a$ and $b$:

$$a + b = 1, \qquad ab = -6$$

So $a$ and $b$ are the roots of $t^2 – t – 6 = 0$, that is $(t-3)(t+2) = 0$:

$$t = 3 \text{ or } t = -2$$

Two assignments, two lines.

$a = 3$, $b = -2$:

$$\frac x3 + \frac{y}{-2} = 1 \quad\Longrightarrow\quad 2x – 3y = 6$$

$a = -2$, $b = 3$:

$$\frac{x}{-2} + \frac y3 = 1 \quad\Longrightarrow\quad -3x + 2y = 6$$

Question 3

What are the points on the $y$-axis whose distance from the line $\dfrac x3 + \dfrac y4 = 1$ is $4$ units?

Solution. Clearing fractions, the line is $4x + 3y – 12 = 0$. A point on the $y$-axis is $(0, c)$, so

$$\frac{\left|4(0) + 3c – 12\right|}{\sqrt{16+9}} = \frac{\left|3c – 12\right|}{5} = 4$$

$$\left|3c – 12\right| = 20$$

$$3c – 12 = 20 \ \Longrightarrow\ c = \frac{32}{3} \qquad\text{or}\qquad 3c – 12 = -20 \ \Longrightarrow\ c = -\frac83$$

Question 4

Find the perpendicular distance from the origin to the line joining the points $(\cos\theta, \sin\theta)$ and $(\cos\phi, \sin\phi)$.

Solution. Both points lie on the unit circle. The line through them, by the two-point form:

$$y – \sin\theta = \frac{\sin\phi – \sin\theta}{\cos\phi – \cos\theta}\left(x – \cos\theta\right)$$

Convert the two differences to products:

$$\sin\phi – \sin\theta = 2\cos\frac{\phi+\theta}{2}\sin\frac{\phi-\theta}{2}, \qquad \cos\phi – \cos\theta = -2\sin\frac{\phi+\theta}{2}\sin\frac{\phi-\theta}{2}$$

so the slope is $-\cot\dfrac{\phi+\theta}{2}$. Writing $\alpha = \dfrac{\phi+\theta}{2}$ and clearing the fraction, the line becomes

$$x\cos\alpha + y\sin\alpha = \cos\theta\cos\alpha + \sin\theta\sin\alpha = \cos(\alpha – \theta)$$

and $\alpha – \theta = \dfrac{\phi – \theta}{2}$. Since $\cos^2\alpha + \sin^2\alpha = 1$, the distance from the origin is simply the size of the right-hand side:

$$d = \left|\cos\frac{\phi – \theta}{2}\right|$$

Question 5

Find the equation of the line parallel to the $y$-axis and drawn through the point of intersection of the lines $x – 7y + 5 = 0$ and $3x + y = 0$.

Solution. Solve the two equations. From the second, $y = -3x$; substituting into the first:

$$x – 7(-3x) + 5 = 0 \quad\Longrightarrow\quad 22x = -5 \quad\Longrightarrow\quad x = -\frac{5}{22}$$

A line parallel to the $y$-axis has the form $x = $ constant, so only the abscissa of the intersection is needed — $y$ never enters the answer:

$$x = -\frac{5}{22}$$

Question 6

Find the equation of a line drawn perpendicular to the line $\dfrac x4 + \dfrac y6 = 1$ through the point where it meets the $y$-axis.

Solution. The line meets the $y$-axis at its $y$-intercept, $(0, 6)$.

Its slope: rewriting as $6x + 4y = 24$, that is $y = -\tfrac32 x + 6$, gives $m = -\tfrac32$. The perpendicular slope is $\tfrac23$.

Through $(0, 6)$:

$$y – 6 = \frac23(x – 0) \quad\Longrightarrow\quad 3y – 18 = 2x \quad\Longrightarrow\quad 2x – 3y + 18 = 0$$

Question 7

Find the area of the triangle formed by the lines $y – x = 0$, $x + y = 0$ and $x – k = 0$.

Solution. Find the three vertices by taking the lines in pairs.

• $y = x$ and $y = -x$ meet at the origin $(0, 0)$.
• $y = x$ and $x = k$ meet at $(k, k)$.
• $y = -x$ and $x = k$ meet at $(k, -k)$.

Two of the vertices share the abscissa $k$, so the side joining them is vertical with length $\left|k – (-k)\right| = 2|k|$, and the perpendicular distance from the origin to that side is $|k|$.

$$\text{area} = \frac12 \times \text{base} \times \text{height} = \frac12 \times 2|k| \times |k| = k^2$$

Question 8

Find the value of $p$ so that the three lines $3x + y – 2 = 0$, $px + 2y – 3 = 0$ and $2x – y – 3 = 0$ may intersect at one point.

Solution. Find where the two lines without $p$ meet, then force the third through it.

Adding $3x + y = 2$ and $2x – y = 3$:

$$5x = 5 \quad\Longrightarrow\quad x = 1, \qquad y = 2 – 3 = -1$$

So the point is $(1, -1)$. Substituting into the second line:

$$p(1) + 2(-1) – 3 = 0 \quad\Longrightarrow\quad p – 5 = 0 \quad\Longrightarrow\quad p = 5$$

Question 9

If three lines whose equations are $y = m_1x + c_1$, $y = m_2x + c_2$ and $y = m_3x + c_3$ are concurrent, show that

$$m_1(c_2 – c_3) + m_2(c_3 – c_1) + m_3(c_1 – c_2) = 0$$

Solution. Let the common point be $(h, k)$, so all three equations hold there:

$$k = m_1h + c_1, \qquad k = m_2h + c_2, \qquad k = m_3h + c_3$$

Subtracting in pairs eliminates $k$:

$$(m_1 – m_2)h = c_2 – c_1, \qquad (m_2 – m_3)h = c_3 – c_2, \qquad (m_3 – m_1)h = c_1 – c_3$$

Now multiply the first by $m_3$, the second by $m_1$ and the third by $m_2$, and add. The left side becomes

$$h\left[m_3(m_1 – m_2) + m_1(m_2 – m_3) + m_2(m_3 – m_1)\right]$$

and the bracket expands to

$$m_1m_3 – m_2m_3 + m_1m_2 – m_1m_3 + m_2m_3 – m_1m_2 = 0$$

so the left side vanishes. The right side is therefore also zero:

$$m_3(c_2 – c_1) + m_1(c_3 – c_2) + m_2(c_1 – c_3) = 0$$

Rearranging the terms by collecting each $m_i$:

$$m_1(c_2 – c_3) + m_2(c_3 – c_1) + m_3(c_1 – c_2) = 0 \qquad \blacksquare$$

Question 10

Find the equations of the lines through the point $(3, 2)$ which make an angle of $45^\circ$ with the line $x – 2y = 3$.

Solution. The given line has slope $\tfrac12$. If the required slope is $m$,

$$\tan 45^\circ = \left|\frac{m – \frac12}{1 + \frac m2}\right| = 1$$

Case $+1$:

$$m – \frac12 = 1 + \frac m2 \quad\Longrightarrow\quad \frac m2 = \frac32 \quad\Longrightarrow\quad m = 3$$

Through $(3,2)$: $y – 2 = 3(x – 3)$, that is $3x – y = 7$.

Case $-1$:

$$m – \frac12 = -1 – \frac m2 \quad\Longrightarrow\quad \frac{3m}{2} = -\frac12 \quad\Longrightarrow\quad m = -\frac13$$

Through $(3,2)$: $y – 2 = -\tfrac13(x-3)$, that is $x + 3y = 9$.

The two answers are perpendicular to each other, as they must be: rotating a line $45^\circ$ each way gives directions $90^\circ$ apart.

Question 11

Find the equation of the line passing through the point of intersection of the lines $4x + 7y – 3 = 0$ and $2x – 3y + 1 = 0$ that has equal intercepts on the axes.

Solution. First the intersection. Multiplying the second equation by $2$ and subtracting from the first:

$$4x + 7y = 3, \qquad 4x – 6y = -2$$

Subtracting: $13y = 5$, so $y = \tfrac{5}{13}$, and then $2x = 3y – 1 = \tfrac{15}{13} – 1 = \tfrac{2}{13}$, giving $x = \tfrac{1}{13}$.

Equal intercepts means the line has the form $x + y = a$. Substituting the intersection point:

$$a = \frac{1}{13} + \frac{5}{13} = \frac{6}{13}$$

$$x + y = \frac{6}{13} \quad\Longrightarrow\quad 13x + 13y = 6$$

Question 12

Show that the equation of the line passing through the origin and making an angle $\theta$ with the line $y = mx + c$ is

$$\frac yx = \frac{m \pm \tan\theta}{1 \mp m\tan\theta}$$

Solution. A line through the origin has equation $y = \mathrm Mx$, so $\mathrm M = \dfrac yx$ is exactly the quantity to be found.

The angle between it and $y = mx + c$ satisfies

$$\tan\theta = \left|\frac{\mathrm M – m}{1 + m\mathrm M}\right|$$

Dropping the modulus gives two cases.

Case $+$:

$$\frac{\mathrm M – m}{1 + m\mathrm M} = \tan\theta \quad\Longrightarrow\quad \mathrm M – m = \tan\theta + m\mathrm M\tan\theta$$

$$\mathrm M\left(1 – m\tan\theta\right) = m + \tan\theta \quad\Longrightarrow\quad \mathrm M = \frac{m + \tan\theta}{1 – m\tan\theta}$$

Case $-$: the identical working with $\tan\theta$ replaced by $-\tan\theta$ gives

$$\mathrm M = \frac{m – \tan\theta}{1 + m\tan\theta}$$

Combining the two into one statement, and writing $\mathrm M = \dfrac yx$:

$$\frac yx = \frac{m \pm \tan\theta}{1 \mp m\tan\theta} \qquad \blacksquare$$

The linked signs — upper with upper, lower with lower — are what encode the two cases in a single line.

Question 13

In what ratio is the line joining $(-1, 1)$ and $(5, 7)$ divided by the line $x + y = 4$?

Solution. Suppose the line divides the segment in the ratio $k : 1$. By the section formula, the point of division is

$$\left(\frac{5k – 1}{k+1},\ \frac{7k + 1}{k+1}\right)$$

This point lies on $x + y = 4$:

$$\frac{5k – 1}{k+1} + \frac{7k+1}{k+1} = 4 \quad\Longrightarrow\quad \frac{12k}{k+1} = 4$$

$$12k = 4k + 4 \quad\Longrightarrow\quad 8k = 4 \quad\Longrightarrow\quad k = \frac12$$

So the ratio is $\tfrac12 : 1$, that is

$$1 : 2$$

The positive value confirms the division is internal — the line genuinely cuts the segment rather than its extension.

Question 14

Find the distance of the line $4x + 7y + 5 = 0$ from the point $(1, 2)$ along the line $2x – y = 0$.

Solution. This is not the perpendicular distance — the direction of travel is prescribed. Note first that $(1,2)$ does lie on $2x – y = 0$, since $2(1) – 2 = 0$ ✓

So find where $2x – y = 0$ meets $4x + 7y + 5 = 0$. Substituting $y = 2x$:

$$4x + 14x + 5 = 0 \quad\Longrightarrow\quad 18x = -5 \quad\Longrightarrow\quad x = -\frac{5}{18}, \qquad y = -\frac{5}{9}$$

Now the distance from $(1, 2)$ to $\left(-\tfrac{5}{18}, -\tfrac59\right)$:

$$\Delta x = 1 + \frac{5}{18} = \frac{23}{18}, \qquad \Delta y = 2 + \frac59 = \frac{23}{9} = \frac{46}{18}$$

$$d = \sqrt{\left(\frac{23}{18}\right)^2 + \left(\frac{46}{18}\right)^2} = \frac{23}{18}\sqrt{1 + 4} = \frac{23\sqrt5}{18}$$

Question 15

Find the direction in which a straight line must be drawn through the point $(-1, 2)$ so that its point of intersection with the line $x + y = 4$ may be at a distance of $3$ units from this point.

Solution. Use the parametric form: a point at distance $r$ from $(-1, 2)$ in the direction $\theta$ is

$$\left(-1 + r\cos\theta,\ 2 + r\sin\theta\right)$$

With $r = 3$, require this point to lie on $x + y = 4$:

$$\left(-1 + 3\cos\theta\right) + \left(2 + 3\sin\theta\right) = 4$$

$$1 + 3\left(\cos\theta + \sin\theta\right) = 4 \quad\Longrightarrow\quad \cos\theta + \sin\theta = 1$$

Squaring: $1 + 2\sin\theta\cos\theta = 1$, so $\sin 2\theta = 0$ and $\theta = 0^\circ$ or $90^\circ$ within one revolution (checking each against the unsquared equation, since squaring can introduce false roots — both survive).

$$\theta = 0^\circ: \text{ the line is parallel to the } x\text{-axis}$$ $$\theta = 90^\circ: \text{ the line is parallel to the } y\text{-axis}$$

Question 16

The hypotenuse of a right-angled triangle has its ends at the points $(1, 3)$ and $(-4, 1)$. Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.

Solution. The two legs are parallel to the axes and meet at the right angle, so the right-angle vertex shares its abscissa with one hypotenuse endpoint and its ordinate with the other. That gives exactly two possibilities.

Vertex at $(1, 1)$: it shares $x = 1$ with $(1,3)$ and $y = 1$ with $(-4,1)$. The legs are

$$x = 1 \quad\text{and}\quad y = 1$$

Vertex at $(-4, 3)$: sharing $x = -4$ with $(-4,1)$ and $y = 3$ with $(1,3)$:

$$x = -4 \quad\text{and}\quad y = 3$$

Question 17

Find the image of the point $(3, 8)$ with respect to the line $x + 3y = 7$, assuming the line to be a plane mirror.

Solution. Let the image be $(h, k)$. Two conditions determine it.

Condition 1 — the midpoint lies on the mirror. The midpoint of $(3,8)$ and $(h,k)$ is $\left(\tfrac{h+3}{2}, \tfrac{k+8}{2}\right)$, and it must satisfy $x + 3y = 7$:

$$\frac{h+3}{2} + 3\cdot\frac{k+8}{2} = 7 \quad\Longrightarrow\quad h + 3 + 3k + 24 = 14 \quad\Longrightarrow\quad h + 3k = -13 \qquad (1)$$

Condition 2 — the join is perpendicular to the mirror. The mirror has slope $-\tfrac13$, so the segment has slope $3$:

$$\frac{k – 8}{h – 3} = 3 \quad\Longrightarrow\quad k – 8 = 3h – 9 \quad\Longrightarrow\quad k = 3h – 1 \qquad (2)$$

Substituting (2) into (1):

$$h + 3(3h – 1) = -13 \quad\Longrightarrow\quad 10h = -10 \quad\Longrightarrow\quad h = -1$$

and then $k = 3(-1) – 1 = -4$.

Question 18

If the lines $y = 3x + 1$ and $2y = x + 3$ are equally inclined to the line $y = mx + 4$, find the value of $m$.

Solution. “Equally inclined” means the line $y = mx + 4$ makes the same angle with each. Writing $m_1 = 3$ and $m_2 = \tfrac12$:

$$\left|\frac{m_1 – m}{1 + m_1m}\right| = \left|\frac{m_2 – m}{1 + m_2m}\right| \quad\Longrightarrow\quad \frac{3 – m}{1 + 3m} = \pm\frac{\frac12 – m}{1 + \frac m2}$$

The sign that gives a genuine solution is the negative one — with $+$ the equation reduces to $m_1 = m_2$, which is false. So

$$\frac{3-m}{1+3m} = -\frac{1 – 2m}{2 + m}$$

Cross-multiplying:

$$(3-m)(2+m) = -(1-2m)(1+3m)$$

$$6 + 3m – 2m – m^2 = -\left(1 + 3m – 2m – 6m^2\right)$$

$$6 + m – m^2 = -1 – m + 6m^2$$

$$7m^2 – 2m – 7 = 0$$

By the quadratic formula:

$$m = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt2}{14} = \frac{1 \pm 5\sqrt2}{7}$$

Question 19

If the sum of the perpendicular distances of a variable point $\mathrm P(x, y)$ from the lines $x + y – 5 = 0$ and $3x – 2y + 7 = 0$ is always $10$, show that $\mathrm P$ must move on a line.

Solution. The two distances are

$$d_1 = \frac{\left|x + y – 5\right|}{\sqrt{1^2 + 1^2}} = \frac{\left|x+y-5\right|}{\sqrt2}, \qquad d_2 = \frac{\left|3x – 2y + 7\right|}{\sqrt{9+4}} = \frac{\left|3x-2y+7\right|}{\sqrt{13}}$$

The condition is

$$\frac{\left|x+y-5\right|}{\sqrt2} + \frac{\left|3x-2y+7\right|}{\sqrt{13}} = 10$$

Within any one region of the plane — that is, on a fixed side of each of the two lines — the signs inside the moduli are constant, so each modulus can be removed with a definite sign. Taking, for instance, both expressions positive:

$$\frac{x+y-5}{\sqrt2} + \frac{3x-2y+7}{\sqrt{13}} = 10$$

Multiplying through by $\sqrt{26}$ and collecting terms:

$$\left(\sqrt{13} + 3\sqrt2\right)x + \left(\sqrt{13} – 2\sqrt2\right)y – \left(5\sqrt{13} – 7\sqrt2 + 10\sqrt{26}\right) = 0$$

This has the form $\mathrm Ax + \mathrm By + \mathrm C = 0$ with constant $\mathrm A$, $\mathrm B$, $\mathrm C$ — a straight line. The other three sign choices give three further linear equations, so in every region $\mathrm P$ moves along a line. $\blacksquare$

Question 20

Find the equation of the line which is equidistant from the parallel lines $9x + 6y – 7 = 0$ and $3x + 2y + 6 = 0$.

Solution. The two lines are parallel — the second is the first with coefficients divided by $3$ — but their coefficients are not yet identical. Scale the second by $3$ first:

$$3x + 2y + 6 = 0 \quad\Longrightarrow\quad 9x + 6y + 18 = 0$$

Now both have $9x + 6y$, and the required line, being equidistant, has the same coefficients with the average constant:

$$c = \frac{-7 + 18}{2} = \frac{11}{2}$$

$$9x + 6y + \frac{11}{2} = 0 \quad\Longrightarrow\quad 18x + 12y + 11 = 0$$

Skipping the scaling step and averaging $-7$ and $6$ directly is the classic error here; the constants are only comparable once the coefficients match.

Question 21

A ray of light passing through the point $(1, 2)$ reflects on the $x$-axis at point $\mathrm A$ and the reflected ray passes through the point $(5, 3)$. Find the coordinates of $\mathrm A$.

Solution. Let $\mathrm A = (a, 0)$. The law of reflection says the angle of incidence equals the angle of reflection, which — measured against the $x$-axis — means the incident and reflected rays have slopes that are negatives of each other.

The neatest route uses the image: reflecting $(1, 2)$ in the $x$-axis gives $(1, -2)$, and the reflected ray, extended backwards, passes through that image. So $\mathrm A$ lies on the straight line joining $(1, -2)$ and $(5, 3)$.

That line has slope $\dfrac{3 – (-2)}{5 – 1} = \dfrac54$, and passes through $(1, -2)$:

$$y + 2 = \frac54(x – 1)$$

Setting $y = 0$ to find where it crosses the $x$-axis:

$$2 = \frac54(x-1) \quad\Longrightarrow\quad x – 1 = \frac85 \quad\Longrightarrow\quad x = \frac{13}{5}$$

Question 22

Prove that the product of the lengths of the perpendiculars drawn from the points $\left(\sqrt{a^2-b^2},\ 0\right)$ and $\left(-\sqrt{a^2-b^2},\ 0\right)$ to the line $\dfrac xa\cos\theta + \dfrac yb\sin\theta = 1$ is $b^2$.

Solution. Write the line as

$$\frac{\cos\theta}{a}x + \frac{\sin\theta}{b}y – 1 = 0$$

and put $c = \sqrt{a^2 – b^2}$ and $\mathrm D = \sqrt{\dfrac{\cos^2\theta}{a^2} + \dfrac{\sin^2\theta}{b^2}}$.

The two distances are

$$p_1 = \frac{\left|\frac{c\cos\theta}{a} – 1\right|}{\mathrm D}, \qquad p_2 = \frac{\left|\frac{-c\cos\theta}{a} – 1\right|}{\mathrm D} = \frac{\left|\frac{c\cos\theta}{a} + 1\right|}{\mathrm D}$$

Their product is a difference of two squares:

$$p_1p_2 = \frac{\left|\frac{c^2\cos^2\theta}{a^2} – 1\right|}{\mathrm D^2}$$

Numerator. Substituting $c^2 = a^2 – b^2$:

$$\frac{(a^2-b^2)\cos^2\theta}{a^2} – 1 = \cos^2\theta – \frac{b^2\cos^2\theta}{a^2} – 1 = -\sin^2\theta – \frac{b^2\cos^2\theta}{a^2}$$

so its modulus is $\dfrac{a^2\sin^2\theta + b^2\cos^2\theta}{a^2}$.

Denominator.

$$\mathrm D^2 = \frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2} = \frac{b^2\cos^2\theta + a^2\sin^2\theta}{a^2b^2}$$

Dividing, the identical bracket cancels:

$$p_1p_2 = \frac{a^2\sin^2\theta + b^2\cos^2\theta}{a^2} \times \frac{a^2b^2}{a^2\sin^2\theta + b^2\cos^2\theta} = b^2 \qquad \blacksquare$$

Question 23

A person standing at the junction (crossing) of two straight paths represented by the equations $2x – 3y + 4 = 0$ and $3x + 4y – 5 = 0$ wants to reach the path whose equation is $6x – 7y + 8 = 0$ in the least time. Find the equation of the path that they should follow.

Solution. “Least time” means the shortest distance, and the shortest route from a point to a line is along the perpendicular. So the required path is the perpendicular from the junction to $6x – 7y + 8 = 0$.

The junction. Solve the two path equations. Multiplying the first by $4$ and the second by $3$:

$$8x – 12y = -16, \qquad 9x + 12y = 15$$

Adding: $17x = -1$, so $x = -\tfrac{1}{17}$. Then from $2x – 3y + 4 = 0$:

$$3y = 2\left(-\frac{1}{17}\right) + 4 = \frac{-2 + 68}{17} = \frac{66}{17} \quad\Longrightarrow\quad y = \frac{22}{17}$$

The slope. The target path $6x – 7y + 8 = 0$ has slope $\tfrac67$, so the perpendicular has slope $-\tfrac76$.

The equation. Through $\left(-\tfrac{1}{17},\ \tfrac{22}{17}\right)$:

$$y – \frac{22}{17} = -\frac76\left(x + \frac{1}{17}\right)$$

Multiplying by $102 = 6 \times 17$:

$$6(17y – 22) = -7(17x + 1)$$

$$102y – 132 = -119x – 7 \quad\Longrightarrow\quad 119x + 102y = 125$$

Common mistakes

• Question 1(a) and (b), swapping the conditions. A line parallel to the $x$-axis has no $x$ term ($\mathrm A = 0$), and one parallel to the $y$-axis has no $y$ term. Checking against $y = 5$ and $x = 5$ settles which is which.
• Question 2, giving only one line. The two intercepts $3$ and $-2$ can be assigned either way round, producing two distinct lines.
• Question 10 and 18, taking only one sign. The angle formula’s modulus always yields two possibilities. In question 10 both are valid; in question 18 the positive case degenerates and must be discarded — but only after being examined.
• Question 13, forgetting to interpret the sign of $k$. A positive ratio means internal division, a negative one external. Here $k = \tfrac12 > 0$, so the segment is genuinely cut.
• Question 14, computing the perpendicular distance. The question specifies travelling along $2x – y = 0$, so the answer is the distance to the intersection point along that line — $\tfrac{23\sqrt5}{18}$, not the shorter perpendicular distance.
• Question 15, forgetting to check the squared solutions. Squaring $\cos\theta + \sin\theta = 1$ can introduce roots that fail the original equation; both $0^\circ$ and $90^\circ$ survive, but $180^\circ$ and $270^\circ$ do not.
• Question 20, averaging the constants without rescaling. $9x + 6y – 7 = 0$ and $3x + 2y + 6 = 0$ must first be written with the same coefficients; averaging $-7$ and $6$ gives a line that is not equidistant at all.
• Question 23, finding the perpendicular from the wrong point or to the wrong line. The person starts at the junction of the first two paths and travels to the third.

Practise next

• Exercise 10.1 — the next chapter, Conic Sections, which begins with circles and uses the distance formula throughout.
• Exercise 9.3 — worth working again alongside questions 3, 14 and 20, since all three are distance-formula problems.